Ví dụ 3: Cho \(\vec{a} = (1; 2)\), \(\vec{b} = (-3; 4)\); \(\vec{c} = (-1; 3)\). Tìm tọa độ của vectơ \(\vec{u}\) biết
a) \(2\vec{u} - 3\vec{a} + \vec{b} = \vec{0}\)
b) \(3\vec{u} + 2\vec{a} + 3\vec{b} = 3\vec{c}\)
Ví dụ 4: Cho \(\vec{a} = (1; 2)\) và \(\vec{b} = (3; 4)\). Tìm độ dài của các vectơ \(\vec{a}\), \(\vec{b}\) và \(2\vec{a} + 3\vec{b}\)
Vd3:
a: \(2\overrightarrow{u}-3\overrightarrow{a}+\overrightarrow{b}=0\)
=>\(\left\{{}\begin{matrix}2x-3\cdot1+\left(-3\right)=0\\2y-3\cdot2+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=6\\2y=4-6=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
=>\(\overrightarrow{u}=\left(3;-1\right)\)
b: \(3\cdot\overrightarrow{u}+2\cdot\overrightarrow{a}+3\cdot\overrightarrow{b}=3\cdot\overrightarrow{c}\)
=>\(\left\{{}\begin{matrix}3x+2\cdot1+3\cdot\left(-3\right)=3\cdot\left(-1\right)\\3y+2\cdot2+3\cdot4=3\cdot3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-7=-3\\3y+20=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-11\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{11}{3}\end{matrix}\right.\)
=>\(\overrightarrow{u}=\left(\dfrac{4}{3};-\dfrac{11}{3}\right)\)
Vd4:
\(\left|\overrightarrow{a}\right|=\sqrt{1^2+2^2}=\sqrt{5}\)
\(\left|\overrightarrow{b}\right|=\sqrt{3^2+4^2}=5\)
\(2\overrightarrow{a}+3\overrightarrow{b}=\left(2\cdot1+3\cdot3;2\cdot2+3\cdot4\right)\)
=>\(2\cdot\overrightarrow{a}+3\cdot\overrightarrow{b}=\left(11;16\right)\)
=>\(\left|2\overrightarrow{a}+3\cdot\overrightarrow{b}\right|=\sqrt{11^2+16^2}=\sqrt{377}\)
Câu 3 :
a) \(2\overrightarrow{u}-3\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{0}\Leftrightarrow\overrightarrow{u}=\dfrac{3\overrightarrow{a}+\overrightarrow{b}}{2}\)
\(\Leftrightarrow\overrightarrow{u}=\dfrac{3\left(1;2\right)-\left(-3;4\right)}{2}=\left(3;1\right)\)
b) \(3\overrightarrow{u}+2\overrightarrow{a}+3\overrightarrow{b}=3\overrightarrow{c}\Leftrightarrow\overrightarrow{u}=\dfrac{3\overrightarrow{c}-2\overrightarrow{a}-3\overrightarrow{b}}{3}\)
\(\Leftrightarrow\overrightarrow{u}=\dfrac{3\left(-1;3\right)-2\left(1;2\right)-3\left(-3;4\right)}{3}=\left(\dfrac{4}{3};-\dfrac{7}{3}\right)\)
Câu 4 :
\(\left|\overrightarrow{a}\right|=\sqrt{1^2+2^2}=\sqrt{5}\)
\(\left|\overrightarrow{b}\right|=\sqrt{3^2+4^2}=5\)
\(2\overrightarrow{a}+3\overrightarrow{b}=2\left(1;2\right)+3\left(3;4\right)=\left(11;16\right)\)
\(\Rightarrow\left|2\overrightarrow{a}+3\overrightarrow{b}\right|=\sqrt{11^2+16^2}=\sqrt{377}\)
